Lesson Solved problems on angles of a polygon

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Solved problems on angles of a polygon

This lesson is a small collection of typical comparatively simple problems on finding angles of a polygon.
The goal of this text is to teach you by examples to make first steps in this area.

Problem 1

Four interior angles of a pentagon are of 107°, 93°, 89° and 92°, correspondingly. Find the measure of the fifth, missing angle.

Solution

The sum of interior angles of a pentagon is (5-2)*180° = 3*180° = 540°.

So, we need to distract the four known angle measures from 540° to get the measure of the missing fifth angle:

540° - (107° + 93° + 89° + 92°) = 159°.

Answer. The missed fifth angle of the pentagon has the measure of 159°.

Problem 2

One angle of a pentagon is 160°. Four of its other angles are congruent. Find the measures of the angles of the pentagon.

Solution

The sum of interior angles of a pentagon is (5-2)*180° = 3*180° = 540°.

So, we need to distract the known angle measure 160° from 540° and divide by 4:

%28540-160%29%2F4

= 95°.

Answer. The four other angles of the pentagon are of 95° each.

Problem 3

The angles of a 6-sided polygon are in the ratio 2:3:4:4:3:2. Find the angles measures.

Solution

The 6 angles of the hexagon are 2x, 3x, 4x, 4x, 3x and 2x, where x is the measure
of the (now unknown) angle, which is the common measure of the given six angles, according to the condition.

Since the sum of interior angles of any hexagon is (6-2)*180° = 4*180° = 720°, you have an equation for the unknown measure x

2x + 3x + 4x + 4x + 3x + 2x = 720, or

18x = 720.

Hence, x = 40, and the common angle measure is 40°.

Then the angles of the hexagon are of 80°, 120°, 150°, 150°, 120° and 80°.

Answer. The angles of the hexagon are of 80°, 120°, 150°, 150°, 120° and 80°.

Problem 4

A polygon has two of its interior angles to be 120° each and others are 150° each.
Calculate the number of sides of the polygon.

Solution

2*120 + 150*(n-2) = 180*(n-2)     <<<---=== The sum of all interior angles of the polygon


240 = 180*(n-2) - 150*(n-2)


240 = 30*(n-2)   ====>  n-2 =  = 8   ====>  n = 8+2 = 10.   Answer.

Problem 5

A polygon has three of its interior angles to be 120° each and others are 150° each.
Calculate the number of sides of the polygon.

Solution

120*3 + (n-3)*150 = (n-2)*180.


Simplify and solve for n.


120*3 + 150n - 3*150 = 180n - 2*180


120*3 - 3*150 + 2*180 = 180n - 150n


270 = 30 n   ====>   n =  = 9.     ANSWER

Problem 6

Two angles of polygon are 120° and 90° and the rest are each 166°. Find the number of the sides.

Solution

120 + 90 + 166*(n-2) = 180*(n-2),

210 + 166n - 332 = 180n - 360,

238 = 14n,

N = 238/14 = 17 sides.

Answer. 17 sides.

My other introductory lessons on finding angles of triangles, parallelograms, quadrilaterals and polygons in this site are
    - Solved problems on supplementary and complementary angles
    - Solved problems on angles of a triangle
    - Solved problems on angles of a parallelogram
    - Solved problems on angles of a quadrilateral
    - Solved problems on missed angle of a polygon
    - Solved problems on angles of a regular polygon
    - OVERVIEW of solved problems on angles of triangles, parallelograms, quadrilaterals and polygons

Use this file/link ALGEBRA-I - YOUR ONLINE TEXTBOOK to navigate over all topics and lessons of the online textbook ALGEBRA-I.

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